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A measure is locally finite if it is finite on all compact sets from the underlying $\sigma$-algebra.

A measure is regular if every measurable set $A\in\Sigma$ can be approximated from above by open measurable sets: $$ \mu(A)\,=\,\operatorname{inf}\,\left\{\,\mu(O)\;\Big{|}\;A\subseteq O\;,\;\;O\in\Sigma\,,\;\,O\;\,\mbox{open} \,\right\}\;\;,\;\; $$ and from below by compact measurable sets: $$ \mu(A)\,=\,\operatorname{sup}\,\left\{\,\mu(C)\;\Big{|}\;C\subseteq A\,,\;\,C\in\Sigma\,,\;\,C\;\,\mbox{compact}\,\right\}\;\;. $$

A Radon measure is a Borel measure that is introduced on the Borel algebra of a Hausdorff topological space, and is regular and locally finite.

Now, my question: $\;\;$is a non-zero Radon measure always positive on non-empty open sets?

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    $\begingroup$ you should specify that $\mu$ be non-zero (otherwise the answer is trivially no) $\endgroup$
    – peek-a-boo
    Jan 17, 2021 at 2:48

3 Answers 3

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Unless I'm overlooking something trivial, the answer is no. Look at the "restriction" of Lebesgue measure $\lambda$ on $\Bbb{R}$ to $[0,1]$, by which I mean $\mu(E):= \lambda(E\cap [0,1])$. Then, $\mu$ is a finite Borel measure on $\Bbb{R}$, hence is regular, but the open set $(10,11)$ has measure $0$ with respect to $\mu$.

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  • $\begingroup$ Thank you. And the Haar measure? $\endgroup$ Jan 17, 2021 at 3:01
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    $\begingroup$ @Michael_1812 I actually don't know much about Haar measures, so I'm not the right person to ask lol (if you want, update your question, and maybe someone else can answer) $\endgroup$
    – peek-a-boo
    Jan 17, 2021 at 3:03
  • $\begingroup$ Thank you. Will do. But first will cogitate on this a bit. $\endgroup$ Jan 17, 2021 at 3:13
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As mentioned in the comments, you should require $\mu \ne 0$. But even then, there are trivial counterexamples.

For example, let $U$ be a non-empty subset of any locally compact Hausdorff space $X$ and $x \notin U$. Then consider the Dirac measure $\delta_x$, which is Radon, and note that $\delta_x(U) = 0.$

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  • $\begingroup$ Yes, that's an eloquent counter-example. Thank you. $\endgroup$ Feb 5 at 16:33
  • $\begingroup$ Could you please answer my question also for the Haar measure? $\endgroup$ Feb 5 at 16:34
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    $\begingroup$ @Michael_1812 For the Haar measure the assertion is true, there is no counterexample. $\endgroup$ Feb 5 at 20:40
  • $\begingroup$ Thank you! Do you happen to know a simple proof or a reference? $\endgroup$ Feb 6 at 4:26
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    $\begingroup$ @Michael_1812 It is in Folland's book "Real analysis" in the chapter about Haar measure. $\endgroup$ Feb 6 at 7:38
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I deeply thank the colleagues who answered my question about the Radon measure.

I think I now can answer the question about the Haar measure.

Suppose there is an open set $\sigma_0$ such that $\mu(\sigma_0)=0$. For an arbitrary open set $\sigma$, consider the infinite union $$ U=\bigcup_{x\in\sigma,\,y\in\sigma_0} g(y,\,x)\,\sigma_0~\,, $$ where $g(y,\,x)$ is the group element sending a point $y\in\sigma_0$ to a point $x\in\sigma$. Since $x$ ranges over the entire $\sigma$, we observe that $$ \sigma\in U~. $$ By definition of topological group, $g(y,\,x)\sigma_0$ is open for $\forall x,\,y$. For a Haar measure, $\mu(\, g(y,\,x)\sigma_0\,)=\mu(\sigma_0)=0$, whence $\mu(U)=0$. By definition of topological space, $U$ is open. We then observe that $\sigma\in U$ and $\mu(U)=0$ together yield $\mu(\sigma)=0$. Thus the existence of one measure-zero open set $\sigma_0$ ensures that all open sets are of zero measure. Among these, is our entire topological space. So $\mu$ is a zero norm, which contradicts the initial assumption.

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